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3 years ago

Determine the molarity of 56.00 grams of aluminum nitrate in 450 milliliters of solution?

Chemistry

1 answer:

SVETLANKA909090 [29]3 years ago

3 0

Answer:

4.62 M

Explanation:

Molarity = moles/volumes (L), so you need to find the moles and the volumes in liters.

Finding the volume is easy because you just have to convert mL to L, so the volume is 0.45 L

Next, find the moles. You can do this by using the molar mass of aluminum to convert the grams to moles. The molar mass of aluminum is 26.98 g/mol.

56 g * (1 mol/26.98 g) = 2.08 mol

Now, divide the moles (2.08) by the volume (.45 L)

Molarity = 4.62 M

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Write Huckel's rule below and determine how many electrons are required to make an aromatic ring with n = 0, 1, and 2.

lisov135 [29]

Answer: The number of electrons for n = 0, 1 and 2 are 2, 6 and 10 respectively.

Explanation:

Huckel's rule is used to determine the aromaticity in a compound. The number of delocalized $\pi-$ electrons are calculated by using the equation:

$\text{Number of delocalized }\pi-\text{ electrons}=4n+2$

where,

n = 0 or any whole number

Calculating the value of electrons for n = 0

Putting values in above equation, we get:

$\text{Number of delocalized }\pi-\text{ electrons}=4(0)+2=2$

Calculating the value of electrons for n = 1

Putting values in above equation, we get:

$\text{Number of delocalized }\pi-\text{ electrons}=4(1)+2=6$

Calculating the value of electrons for n = 2

Putting values in above equation, we get:

$\text{Number of delocalized }\pi-\text{ electrons}=4(2)+2=10$

Hence, the number of electrons for n = 0, 1 and 2 are 2, 6 and 10 respectively.

4 0

3 years ago

Acids and bases can neutralize each other in double displacement reactions. For example, if hydrogen chloride (a strong acid, HC

scoray [572]

Answer:

The answer to your question is HCl + NaOH ⇒ NaCl + H₂O

Explanation:

Data

Double displacement reaction

Balanced chemical reaction

HCl + NaOH ⇒ NaCl + H₂O

Reactants Elements Products

1 Chlorine (Cl) 1

1 Sodium (Na) 1

2 Hydrogen (H) 2

1 Oxygen (0) 1

As we can see, the reaction is balanced and the coefficients of all reactants and products are 1, but the number is not written in a balanced reaction.

8 0

3 years ago

Write the balanced chemical equations for the decomposition of solid calcium hydroxide into solid calcium (ii) oxide (lime) and

Lesechka [4]

Calcium oxide (CaO) or lime in solid form can be prepared from the decomposition of calcium hydroxide {Ca $(OH)_{2}$ in to lime (CaO) and water ( $H_{2}O$ at high temperature. The reaction is an endothermic reaction. That is heat is absorbed in this reaction process. One mole of calcium hydroxide decomposed into one mole of calcium oxide and one mole of water. The balanced reaction can be shown as-CaC $O_{3}$ (solid) → CaO (solid) + $H_{2}O$ (liquid). The heat of the reaction is (+) 63.7 kJ/mole of CaO.

3 0

3 years ago

It takes 11.2 kj of energy to raise the temperature of 145 g of benzene from 22.0°c to 67.0°c. what is the specific heat of benz

Cloud [144]

We can use the heat equation,
Q = mcΔT

where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J g⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).
Q = 11.2 kJ = 11200 J
m = 145 g
c = ?
ΔT = (67 - 22) °C = 45 °C
By applying the formula,
11200 J = 145 g x c x 45 °C
c = 1.72 J g⁻¹ °C⁻¹

Hence, specific heat of benzene is 1.72 J g⁻¹ °C⁻¹.

7 0

3 years ago

What is the unknown metal if the temperature of a beaker of 100ml of water was raised 17c to 19 c when 21 grams of the metal at

horrorfan [7]

Answer:

The metal has a heat capacity of 0.385 J/g°C

This metal is copper.

Explanation:

Step 1: Data given

Mass of the metal = 21 grams

Volume of water = 100 mL

⇒ mass of water = density * volume = 1g/mL * 100 mL = 100 grams

Initial temperature of metal = 122.5 °C

Initial temperature of water = 17°C

Final temperature of water and the metal = 19 °C

Heat capacity of water = 4.184 J/g°C

Step 2: Calculate the specific heat capacity

Heat lost by the metal = heat won by water

Qmetal = -Qwater

Q = m*c*ΔT

m(metal) * c(metal) * ΔT(metal) = - m(water) * c(water) * ΔT(water)

21 grams * c(metal) *(19-122.5) = -100 * 4.184 * (19-17)

-2173.5 *c(metal) = -836.8

c(metal) = 0.385 J/g°C

The metal has a heat capacity of 0.385 J/g°C

This metal is copper.

4 0

3 years ago